399.除法求值

给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。

另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。

返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。如果问题中出现了给定的已知条件中没有出现的字符串，也需要用 -1.0 替代这个答案。

注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。

示例 1：

输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]

示例 2：

输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]

示例 3：

输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]

提示：

• 1 <= equations.length <= 20
• equations[i].length == 2
• 1 <= Ai.length, Bi.length <= 5
• values.length == equations.length
• 0.0 < values[i] <= 20.0
• 1 <= queries.length <= 20
• queries[i].length == 2
• 1 <= Cj.length, Dj.length <= 5
• Ai, Bi, Cj, Dj 由小写英文字母与数字组成

方法一：bfs

时间复杂度：O(ML+Q·(L+M))

空间复杂度：O(NL+M)

/**
 * @param {string[][]} equations
 * @param {number[]} values
 * @param {string[][]} queries
 * @return {number[]}
 */
var calcEquation = function(equations, values, queries) {
    let nvars = 0;
    const variables = new Map();

    const n = equations.length;
    for (let i = 0; i < n; i++) {
        if (!variables.has(equations[i][0])) {
            variables.set(equations[i][0], nvars++);
        }
        if (!variables.has(equations[i][1])) {
            variables.set(equations[i][1], nvars++);
        }
    }

    // 对于每个点，存储其直接连接到的所有点及对应的权值
    const edges = new Array(nvars).fill(0);
    for (let i = 0; i < nvars; i++) {
        edges[i] = [];
    }
    for (let i = 0; i < n; i++) {
        const va = variables.get(equations[i][0]), vb = variables.get(equations[i][1]);
        edges[va].push([vb, values[i]]);
        edges[vb].push([va, 1.0 / values[i]]);
    }

    const queriesCount = queries.length;
    const ret = [];
    for (let i = 0; i < queriesCount; i++) {
        const query = queries[i];
        let result = -1.0;
        if (variables.has(query[0]) && variables.has(query[1])) {
            const ia = variables.get(query[0]), ib = variables.get(query[1]);
            if (ia === ib) {
                result = 1.0;
            } else {
                const points = [];
                points.push(ia);
                const ratios = new Array(nvars).fill(-1.0);
                ratios[ia] = 1.0;

                while (points.length && ratios[ib] < 0) {
                    const x = points.pop();
                    for (const [y, val] of edges[x]) {
                        if (ratios[y] < 0) {
                            ratios[y] = ratios[x] * val;
                            points.push(y);
                        }
                    }
                }
                result = ratios[ib];
            }
        }
        ret[i] = result;
    }
    return ret;
};